One more time.
It was a WORST CASE calculation which placed an upper bound on the possible
loss
from operating into a 1.7:1 SWR as opposed to a 1:1 SWR.
In practice most of the power is eventually radiated.
sheesh!
H.


"Cecil Moore" wrote in message
...
Fred W4JLE wrote:
I would disagree with your statement about SWR being absorbed in the
source.
The reflected wave is rereflected and aside from losses caused by the 2
way
trip is reradiated.

Reflected current flowing into the final amp can superpose in
phase with the forward current and, without protection circuitry,
cause the final amp to exceed its dissipation rating.

Another possibility is when the reflected voltage superposes
in phase with the forward voltage, and without protection
circuitry, exceeds the voltage rating of the final.

If all reflected power was always re-reflected, there would
be no need for protection circuitry. The generated power is
*defined* as the forward power minus the reflected power. That
does NOT mean that the reflected power is 100% re-reflected.
--
73, Cecil http://www.qsl.net/w5dxp

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Ivory Soap. Excellent!
73
H.


"Fred W4JLE" wrote in message
...
You have created an all encompassing case from the original SWR of 1.7:1
No
tuner needed. Under these conditions, the amount of reflected energy
radiated approaches Ivory Soap.


"Richard Clark" wrote in message
...
On Sat, 4 Jun 2005 22:02:48 -0400, "Fred W4JLE"
wrote:

I would disagree with your statement about SWR being absorbed in the
source.
The reflected wave is rereflected

Hi Fred,

I've seen this thesis offered before. I generally ask, since this is
exactly the same thesis offered for a conjugate match offered by a
tuner, and a tuner is used for this very purpose (rereflecting the
mismatch), if the Transmitter already does it - What is the Tuner for?

Why do you use a tuner? What is a match? Why would anyone seek to
match a Transmitter to its load?

73's
Richard Clark, KB7QHC

In TV broadcasting reflections from the antenna back to the transmitter will
be reflected by the transmitter to the antenna and the signal will be
rebroadcast albeit at somewhat less power. Then depending on the length of
transmission line the viewer may see ghosting.
In audio I don't know why and I have run my Collins 30S-1 into ladder line
with a 14 to SWR with no one except me knowing!

--

73
Hank WD5JFR
"Richard Clark" wrote in message
...
On Sun, 5 Jun 2005 10:02:17 -0400, "Fred W4JLE"
wrote:

You have created an all encompassing case from the original SWR of 1.7:1

Thanx Fred,

No one has ever been able to answer when the Transmitter rereflects
energy, why they need a tuner.

73's
Richard Clark, KB7QHC

H. Adam Stevens, NQ5H wrote:
Roy
Worst case.
Get it?
H.

No, sorry, I don't. The result will be as I posted in the best case,
worst case, and typical case.

What conditions would cause it to be different?

Roy Lewallen, W7EL

H. Adam Stevens, NQ5H wrote:
One more time.
It was a WORST CASE calculation which placed an upper bound on the possible
loss
from operating into a 1.7:1 SWR as opposed to a 1:1 SWR.
In practice most of the power is eventually radiated.
sheesh!
H.

Please give us the conditions (cable Z0, load Z, or whatever you need)
that cause this "worst case" to occur.

Roy Lewallen, W7EL

Fred W4JLE wrote:
Cecil, for every question, there are any number of answers if you refashion
the question. In this case the SWR was defined as !.7:1

!.7:1???? I must admit, that's a new one on me. :-)

My statements were general, and apply to any SWR.
--
73, Cecil http://www.qsl.net/w5dxp

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H. Adam Stevens, NQ5H wrote:

One more time.
It was a WORST CASE calculation which placed an upper bound on the possible
loss from operating into a 1.7:1 SWR as opposed to a 1:1 SWR.
In practice most of the power is eventually radiated.

As Roy, W7EL, previously indicated:

Power reflection coefficient:
rho^2 = [(SWR-1)/(SWR+1)]^2 = 0.067 = Pref/Pfor

1-rho^2 = 0.933 = 93.3% forward power delivered to the load

93.3 watts is 0.3 dB down from 100 watts.

The log of the ratio of two SWRs doesn't seem to have much
meaning.
--
73, Cecil http://www.qsl.net/w5dxp

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Good Lord Roy, I thought you knew better.

If the match at the load is not perfect, energy is refleced back to the
source, are you with me so far?

I can easily build a source that absorbs all the reflected power: A zero
impedance source in series with a resistor that matches the transmission
line impedance.

73
H.


"Roy Lewallen" wrote in message
...
H. Adam Stevens, NQ5H wrote:
Roy
Worst case.
Get it?
H.

No, sorry, I don't. The result will be as I posted in the best case,
worst case, and typical case.

What conditions would cause it to be different?

Roy Lewallen, W7EL

"Roy Lewallen" wrote in message
...
H. Adam Stevens, NQ5H wrote:
One more time.
It was a WORST CASE calculation which placed an upper bound on the
possible
loss
from operating into a 1.7:1 SWR as opposed to a 1:1 SWR.
In practice most of the power is eventually radiated.
sheesh!
H.

Please give us the conditions (cable Z0, load Z, or whatever you need)
that cause this "worst case" to occur.

Roy Lewallen, W7EL

See other post.

It's called db, Cecil.


"Cecil Moore" wrote in message
...
H. Adam Stevens, NQ5H wrote:

One more time.
It was a WORST CASE calculation which placed an upper bound on the
possible
loss from operating into a 1.7:1 SWR as opposed to a 1:1 SWR.
In practice most of the power is eventually radiated.

As Roy, W7EL, previously indicated:

Power reflection coefficient:
rho^2 = [(SWR-1)/(SWR+1)]^2 = 0.067 = Pref/Pfor

1-rho^2 = 0.933 = 93.3% forward power delivered to the load

93.3 watts is 0.3 dB down from 100 watts.

The log of the ratio of two SWRs doesn't seem to have much
meaning.
--
73, Cecil http://www.qsl.net/w5dxp

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